Question

Ammonia under pressure of 15 atm at $27{}^\circ $C is heated to $347{}^\circ $C in a closed vessel in the presence of a catalyst. Under the conditions, $N{{H}_{3}}$is partially decomposed according to the equation, $2N{{H}_{3}}\rightleftharpoons {{N}_{2}}+3{{H}_{2}}$. The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. calculate the percentage of actually decomposed?
A. 65%
B. 61.3%
C. 63.5%
D. 64%

Answer 1

Hint: The problem can be solved by defining the ideal gas law which can be defined as the equation of state of hypothetical ideal gas and it is a good approximation of behavior of many gases under many conditions but it also suffers from many limitations.

Complete step-by-step answer: Equation given is:
$2N{{H}_{3}}\rightleftharpoons {{N}_{2}}+3{{H}_{2}}$

1	0	0
2-2x	x	3x

According to the question we can say that total number of moles at initial stage is 2 and
Total number of moles at equilibrium = 2+2x
Now by applying ideal gas law equation which is PV = nRT
Which suggests that $V=\dfrac{nRT}{P}$
Here $\dfrac{{{n}_{1}}R{{T}_{1}}}{{{P}_{1}}}=\dfrac{{{n}_{2}}R{{T}_{2}}}{{{P}_{2}}}$(As volume is constant)
By putting the values given in question, ${{P}_{1}}=15atm$(Before decomposition), ${{P}_{2}}=50atm$(after decomposition), ${{T}_{1}}=27{}^\circ C=300K$(before decomposition), ${{T}_{2}}=347{}^\circ C=620K$(after decomposition) , ${{n}_{1}}=2$and ${{n}_{2}}=2+2x$.
Now by putting all the values in the equation we find the value of x will be 0.6129 whose percentage can be calculated by multiplying with 100 which would come to be 61.29%.

Thus we can say that option B is the correct answer.

Note: Mole is represented by the symbol mol which is defined as the unit of measurement for the amount of substance in the SI units where SI stands for International System of Units. It can also be defined as exactly $6.022140796\times {{10}^{23}}$particles which may be atoms, molecules, ions or electrons.